chap15-232-253

Figure 1512 equivalent circuit of circulating current

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Unformatted text preview: ystem 15.8.1 Stability It is not feasible to calculate the spill current that may occur, but, fortunately, this is not necessary; an alternative approach provides both the necessary information and the technique required to obtain a high performance. Network Protection &amp; Automation Guide The current transformers are replaced in the diagram by ideal current transformers feeding an equivalent circuit that represents the magnetising losses and secondary winding resistance, and also the resistance of the connecting leads. These circuits can then be interconnected as shown, with a relay connected to the junction points to form the complete equivalent circuit. Saturation has the effect of lowering the exciting impedance, and is assumed to take place severely in current transformer H until, at the limit, the shunt impedance becomes zero and the CT can produce no output. This condition is represented by a short circuit, shown in broken line, across the exciting impedance. It should be noted that this is not the equivalent of a physical short circuit, since it is behind the winding resistance . Applying the Thévenin method of solution, the voltage developed across the relay will be given by: IR= Vf R R + R LH + R CTH ...Equation 15.1 The current through the relay is given by: = I f ( R LH + R CTH ) R R + R LH + R CTH ...Equation 15.2 If RR is small, IR will approximate to IF, which is unacceptable. On the other hand, if RR is large IR is reduced. Equation 15.2 can be written, with little error, as follows: • 241 • Busbar P rotection The incidence of fault current with an initial unilateral transient component causes an abnormal build-up of flux in a current transformer, as described in Section 6.4.10. When through-fault current traverses a zone protected by a differential system, the transient flux produced in the current transformers is not detrimental as long as it remains within the substantially linear range of the magnetising characteristic. With fault current of appreciable magnitude and long transient time constant, the flux density will pass into the saturated region of the characteristic; this will not in itself produce a spill output from a pair of balancing current transformers provided that these are identical and equally burdened. A group of current transformers, though they may be of the same design, will not be completely identical, but a more important factor is inequality of burden. In the case of a differential system for a busbar, an external fault may be fed through a single circuit, the current being supplied to the busbar through all other circuits. The faulted circuit is many times more heavily loaded than the others and the corresponding current transformers are likely to be heavily saturated, while those of the other circuits are not. Severe unbalance is therefore probable, which, with a relay of normal burden, could exceed any acceptable current setting. For this reason such systems were at one time always provided with a time delay. This practice is, however, no longer acceptable. • 15 • IR I Vf = = RR f (R LH RL + RCT = lead + CT winding resistance + R CTH ) K RR … Equation 15.3 or alternatively: I RRR = V f = I f ( R LH + R CTH ) It remains to be shown that the setting chosen is suitable. …Equation 15.4 It is clear that, by increasing RR, the spill current IR can be reduced below any specified relay setting. RR is frequently increased by the addition of a series-connected resistor which is known as the stabilising resistor. Busbar P rotection It can also be seen from Equation 15.4 that it is only the voltage drop in the relay circuit at setting current that is important. The relay can be designed as a voltage measuring device consuming negligible current; and provided its setting voltage exceeds the value Vf of Equation 15.4, the system will be stable. In fact, the setting voltage need not exceed Vf, since the derivation of Equation 15.4 involves an extreme condition of unbalance between the G and H current transformers that is not completely realised. So a safety margin is built-in if the voltage setting is made equal to Vf. • 15 • = factor depending on relay design (range 0.7 - 2.0) The current transformers will have an excitation curve which has not so far been related to the relay setting voltage, the latter being equal to the maximum nominal voltage drop across the lead loop and the CT secondary winding resistance, with the maximum secondary fault current flowing through them. Under in-zone fault conditions it is necessary for the current transformers to produce sufficient output to operate the relay. This will be achieved provided the CT knee-point voltage exceeds the relay setting. In order to cater for errors, it is usual to specify that the current transformers should have a knee-point e.m.f. of at least twice the necessary setting voltage; a higher multiple is of advantage in ensuring a high speed of operation. It is necessary to realise that the value of If to be inserted i...
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